Heine-Cantor theorem

Let $𝑓 : 𝑋 \to 𝑌$ be a continuous function between metric spaces and $𝑋$ be a compact. Then $𝑓$ is uniformly continuous. anal

Proof

Let $(𝑋, 𝑑 𝑋)$ and $(𝑌, 𝑑 𝑌)$ be a metric spaces with $𝑋$ compact, with $𝑓 : 𝑋 \to 𝑌$ a continuous mapping between them. Let $𝜖 > 0$ . By continuity, for every $𝑥 \in 𝑋$ there exists $𝛿 𝑥 (𝜖) > 0$ such that $𝑓 (𝑦) \in B 𝜖 / 2 (𝑓 (𝑥))$ for all $𝑦 \in B 𝛿 𝑥 (𝜖) / 2 (𝑥)$ . Then the balls ${B 𝛿 𝑥 (𝜖) / 2 (𝑥) : 𝑥 \in 𝑋}$ form a open cover of $𝑋$ , so by compactness there must exist a finite set of points $(𝑥 𝑖) 𝑛 𝑖 = 1$ whose balls cover the space, i.e. ${B 𝛿 𝑥 𝑖 (𝜖) / 2 (𝑥 𝑖)} 𝑛 𝑖 = 1$ is a finite subcover. Then there exists $𝛿 (𝜖) = m i n {12 𝛿 𝑥 𝑖 (𝜖)} 𝑛 𝑖 = 1$ since it is the minimum of finitely many positive real numbers.

Now we will show that $𝛿 (𝜖)$ meets the requirements for Uniform continuity. Let $𝑥, 𝑦 \in 𝑋$ such that $𝑑 𝑋 (𝑥, 𝑦) < 𝛿 (𝜖)$ . Then $𝑥 \in B 𝛿 𝑥 𝑖 (𝜖) / 2 (𝑥 𝑖)$ for some $0 \leq 𝑖 \leq 𝑛$ . Then by the triangle inequality
$𝑑 (𝑥 𝑖, 𝑦) \leq 𝑑 (𝑥 𝑖, 𝑥) + 𝑑 (𝑥, 𝑦) < 12 𝛿 𝑥 𝑖 (𝜖) + 𝛿 (𝜖) \leq 𝛿 𝑥 𝑖 (𝜖)$
Therefore $𝑥, 𝑦 \in B 𝛿 𝑥 𝑖 (𝜖) (𝑥 𝑖)$ and thus by the original definition of $𝛿 𝑥 𝑖 (𝜖)$ it follows $𝑓 (𝑥), 𝑓 (𝑦) \in B 𝜖 / 2 (𝑓 (𝑥 𝑖))$ . Thus $𝑑 (𝑓 (𝑥), 𝑓 (𝑦)) < 𝜖$ as required.

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Heine-Cantor theorem

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